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NASA is conducting an experiment to find out the fraction of people who black out at...

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. Step 2 of 2 : Suppose a sample of 1373 people is drawn. Of these people, 576 passed out. Using the data, construct the 98% confidence interval for the population proportion of people who black out at G forces greater than 6. Round your answers to three decimal places.

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Answer #1

Solution :

Given that,

n = 1373

x = 576

Point estimate = sample proportion = = x / n =0.420

q = 1 - = 0.58

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.326 * (((0.420 * 0.58) / 1373)

= 0.031

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.420 - 0.031 < p < 0.420 + 0.031

0.389 < p < 0.451

( 0.389 , 0.451)

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